AI 中文总结
研究代数半域的双-UFS正猜想，通过对有限生成代数正半域在循环情形、二次与高次情形分别论证，结合Perron-Frobenius论证及复半域约化定理，证明了有限生成代数双-UFS半域同构于\(\mathbb{N}_0\)。
AI 中文摘要
如果一个半域的加法幺半群和非零乘法幺半群都是唯一分解幺半群，那么这个半域就称为双-UFS。双-UFS正猜想预测，具有此性质的唯一正半域是非负整数。我们证明了有限生成代数正半域的这个猜想。在循环情形下，对于每个正代数数\(\alpha\)，半域\(\mathbb{N}_0[\alpha]\)是双-UFS当且仅当\(\alpha\in\mathbb{N}\)，即\(\mathbb{N}_0[\alpha]=\mathbb{N}_0\)。证明将二次情形（通过分析大于\(1\)的最小加法原子排除了仅有的例子\(\mathbb{N}_0[\sqrt 2]\)和\(\mathbb{N}_0[(1+\sqrt 5)/2]\)）与高次情形（通过显式乘法恒等式迫使极小多项式为不可能的形式）分开。接着给出一个Perron-Frobenius论证表明，如果\(\alpha_1,\ldots,\alpha_n\)是正代数数且\(\mathbb{N}_0[\alpha_1,\ldots,\alpha_n]\)是双-UFS，那么这个半域就是\(\mathbb{N}_0\)。最后，证明了一个关于复半域的约化定理：\(\mathbb{C}\)中每个具有有限多个加法原子的双-UFS子半域都有一个同构实现为正半域。因此，\(\mathbb{C}\)上每个有限生成代数双-UFS半域都同构于\(\mathbb{N}_0\)。
英文摘要
A semidomain is called bi-UFS if both its additive monoid and its nonzero multiplicative monoid are unique factorization monoids. The Bi-UFS Positive Conjecture predicts that the only positive semidomain with this property is the nonnegative integers. We prove this conjecture for finitely generated algebraic positive semidomains. In the cyclic case, we show that for every positive algebraic number $α$, the semidomain $\mathbb{N}_0[α]$ is bi-UFS if and only if $α\in \mathbb{N}$, equivalently $\mathbb{N}_0[α]=\mathbb{N}_0$. The proof separates the quadratic case, where an analysis of the least additive atom larger than $1$ leaves only the examples $\mathbb{N}_0[\sqrt 2]$ and $\mathbb{N}_0[(1+\sqrt 5)/2]$ to exclude, from the higher-degree case, where explicit multiplicative identities force the minimal polynomial into impossible forms. We then give a Perron-Frobenius argument showing that if $α_1,\ldots,α_n$ are positive algebraic numbers and $\mathbb{N}_0[α_1,\ldots,α_n]$ is bi-UFS then this semidomain is $\mathbb{N}_0$. Finally, we prove a reduction theorem for complex semidomains: every bi-UFS subsemidomain of $\mathbb{C}$ with finitely many additive atoms admits an isomorphic realization as a positive semidomain. Consequently, every finitely generated algebraic bi-UFS semidomain over $\mathbb{C}$ is isomorphic to $\mathbb{N}_0$.
CommentsWe made a few minor edits throughout the document
